Calculate the following limit. $$ \lim_{x\to 0} x\left\lfloor\frac{1}{x}\right\rfloor $$ Where $\left\lfloor x \right\rfloor$ represents greatest integer function or floor function, i.e greatest integer less than or equal to $x$.
Thanks.
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6$\begingroup$ Are the square brackets just brackets, or do they refer to something like the integer part, floor or ceiling? Please clarify. $\endgroup$ – Clive Newstead Mar 29 '13 at 18:10
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1$\begingroup$ @CliveNewstead I edited to make it the floor function. That's just a guess, but it often what $[x]$ means, and I doubt OP just means them as parentheses. $\endgroup$ – Thomas Andrews Mar 29 '13 at 18:12
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In case $[x]$ is intended to be an integer part of $x$, you have $$ \lim_{x\to 0}x\left[\frac1x\right] = \lim_{y\to\infty}\frac{[y]}{y} = \lim_{y\to\infty}\frac{y -\{y\}}y = 1. $$
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If $\frac1x=n+y$ where $n$ is any integer and $ 0\le y<1,\implies \left[\frac1x\right]=n$
So, $x \left[\frac1x\right]=\frac n{n+y}=\frac1{1+\frac yn}$
As $x\to 0$ and $ 0\le y<1, n\to\infty\implies \lim_{x\to 0}x \left[\frac1x\right]=\lim_{n\to\infty}\frac1{1+\frac yn}=1$
answered Mar 29 '13 at 18:12
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Consider $x \in (1/(n+1),1/n]$. We then have $\dfrac1x \in [n,n+1)$. Hence, $\left\lfloor \dfrac1x \right\rfloor = n$. Hence, we have $$x \left\lfloor \dfrac1x \right\rfloor \in \left(\dfrac{n}{n+1},1\right]$$ Argue similarly, for $x \to 0^{-}$. Now use the above to show that $$\lim_{x \to 0} x \left\lfloor \dfrac1x \right\rfloor = 1$$
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If $[x]$ means floor (usually noted $\lfloor x \rfloor$), you have that $\lfloor x \rfloor = -1$ if $-1 \le x < 0$, while $\lfloor x \rfloor = 0$ when $0 \le x < 1$. The limit can't exist.
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$\begingroup$ The limit does exist. The interested quantity is $\lfloor 1/x \rfloor$ and not $\lfloor x \rfloor$. $\endgroup$ – user17762 Mar 29 '13 at 18:17
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$\begingroup$ The math rendering here doesn't allow to see if it is $\lfloor 1 / x \rfloor$ or $1 / \lfloor x \rfloor$... $\endgroup$ – vonbrand Mar 29 '13 at 18:18
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Use one basic inequality concerning integer part $$\frac{1}{x}\leq [\frac{1}{x}]\leq \frac{1}{x}+1$$
\begin{align} 1 \leq x[\frac{1}{x}]\leq 1+x \,\mathrm{if}\, x>0\\ 1\geq x[\frac{1}{x}]\geq 1+x \,\mathrm{if}\, x<0 \end{align} So by squeeze theorem $$\lim _{n\rightarrow 0}x[\frac{1}{x}]=0$$
